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rodikova [14]
3 years ago
12

If you wanted to discover the youngest stars you could find in some grouping of stars in the Galaxy, which type of star group wo

uld be the best to search?
Physics
1 answer:
Vilka [71]3 years ago
7 0

Answer:

Associations Clusters

Explanation:

There is three basic type of clusters, globular clusters, open clusters, associations clusters. Associations cluster consist of younger stars that is younger than globular and open clusters stars.

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The engine oil at 150 degree Celsius is cooled to 80 degree Celsius in a parallel flow heat exchanger by water entering at 25 de
Setler [38]

Answer:

Explanation:

Attach is the solution

6 0
3 years ago
A rectangular loop with dimensions 4.20 cm by 9.50 cm carries current I. The current in the loop produces a magnetic field at th
tiny-mole [99]

Answer:

I=1.48 A

Explanation:

Given that

B=3.1 x 10⁻5 T

b= 4.2 cm

l= 9.5 cm

The relationship for magnetic field  and current given as

B=\dfrac{2\mu _oI}{\pi}D

Where

D=\dfrac{\sqrt{l^2+b^2}}{lb}

By putting the values

D=\dfrac{\sqrt{l^2+b^2}}{lb}

D=\dfrac{\sqrt{0.042^2+0.095^2}}{0.042\times 0.095}

D=26.03 m⁻¹

B=\dfrac{2\mu _oI}{\pi}D

3.1\times 10^{-5}=\dfrac{2\times 4\times \pi \times 10^{-7} I}{\pi}\times 26.03

I=\dfrac{3.1\times 10^{-5}}{{2\times 4\times 10^{-7} }\times 26.03}

I=1.48 A

8 0
3 years ago
When a guitar string plays the note "a," the string vibrates at 440 hz ?
Rama09 [41]
Yes, that's correct. The note "A" (which is used to tune the other strings of the guitar) corresponds to a frequency of 440 Hz.
8 0
3 years ago
A length of copper wire carries a current of 14 A, uniformly distributed through its cross section. The wire diameter is 2.5 mm,
gavmur [86]

Answer:

a. ρ_\beta=1.996J/m^3

b. U_E=9.445x10^{-15} J/m^3

Explanation:

a. To find the density of magnetic field given use the gauss law and the equation:

i=14A, d=2.5mm, R=3.3Ω, l=1 km, E_o=8.85x10^{-12}F/m, u_o=4*x10^{-7}H/m

ρ_\beta=\frac{\beta^2}{2*u_o}

ρ_\beta=\frac{1}{2*u_o}*(\frac{u_o*i^2}{2\pi *r})^2

ρ_\beta=\frac{u_o*i^2}{8\pi*r}=\frac{4\pi *10^{-7}H/m*(14A)^2}{8\pi*(1.25x10^{-3}m)^2}

ρ_\beta=1.996J/m^3

b. The electric field can be find using the equation:

U_E=\frac{1}{2}*E_o*E^2

E=(\frac{i*R}{l})^2

U_E=\frac{1}{2}*8.85x10^{-12}*(\frac{14A*3.3}{1000m})^2

U_E=9.445x10^{-15} J/m^3

4 0
3 years ago
A concave spherical mirror has a radius of curvature of magnitude 20.0 cm. (a) Find the location of the image for object distanc
Tems11 [23]

Answer:

Answered

Explanation:

The radius of curvature of the mirror R = 20 cm

then the focal length f = R/2 = 10 cm

(a) From mirror formula

 1/f = 1/di + /1do

then the image distance

    di = fd_o / d_o - f

  = (10)(40) / 40-10

 = 30.76 cm

since the image distance is positive so the image is real

ii) when the object distance d_0=20 cm

 di = 10×20/ 20-10

= 20

Hence, the image must be  real

iii)when the object distance d_0 = 10

di = 10×10 / 10-10 =  ∞ (infinite)

the image will be formed at ∞

here also image will be real but diminished.

7 0
3 years ago
Read 2 more answers
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